# Homework Help: Proof by induction an expression

1. Apr 28, 2012

### xeon123

I'm trying to prove by induction the expression:

$\sum_{i=1}^{n-1}(n-i)=\frac{n(n-1)}{2}$

For the base case, n=2, S(2)=$\frac{2(2-1)}{2}=1$

For S(n+1)=$\frac{(n+1)((n+1)-1)}{2}$ I have:

S(n+1) = $\frac{n(n-1)}{2}$ + (n+1) <--- Is this correct?

I don't know what is the term for n+1. Any help?

Last edited: Apr 28, 2012
2. Apr 28, 2012

### HallsofIvy

There appears to be something missing here! What is being summed?

No one can tell you until you tell us what is being summed.

3. Apr 28, 2012

### xeon123

I corrected in the first message now.

4. Apr 28, 2012

### xeon123

I solved. Here's the solution. Can you check it if it's right?

For n=2, I got
0+1=$\frac{2(2-1)}{2}$

For n=3,
0+1+2=$\frac{3(3-1)}{2}$

So, for n, I got
0+1+2+3+...+n-1=$\frac{n(n-1)}{2}$

For n+1 I got
0+1+2+3+...+n-1+n=$\frac{(n+1)(n+1-1)}{2}=\frac{(n+1)(n)}{2}$

For the LHS

LHS=$\frac{n(n-1)}{2}+n=\frac{n(n-1)+2n}{2}=\frac{n^2-n+2n}{2}=\frac{n^2+n}{2}$

For the RHS
RHS=$\frac{(n+1)(n)}{2}=\frac{n^2+n}{2}$

So, the LHS = RHS, proofing that $\frac{n(n-1)}{2}$ is correct for any n.

Is this solution correct?

5. Apr 28, 2012

### HallsofIvy

What you have written is basically a "proof by induction" and, yes, it is correct.

Actually, the first thing I would do is write that as
$$\sum_{i= 1}^{n- 1} n- 1= \sum_{j= 0}^n j$$
where I have taken j= i- 1.

If I really didn't want to do that, I might note that
$$\sum_{i=1}^{n-1}= n\sum_{i= 1}^n 1- \sum_{i=1}^{n- 1}i$$